A098993 - OEIS

%I #8 Mar 14 2015 16:16:38

%S 5,25,39,50,218,403,403,403,403,1597,2190,2820,6050,8577,12423,27325,

%T 34075,37088,37088,43795,43795,43795,87594,87594,87594,87594,87594,

%U 169160,169160,169160,169160,169160,169160,1884817,1884817,1884817

%N Let h be the smallest value for which h, h+1, ..., h+n-1 are all lengths of hypotenuses of Pythagorean triangles. Then a(n)=h.

%C "We can also prove (this is more difficult) that for an arbitrary natural number m there exist m Pythagorean triangles the hypotenuses of which are given by successive natural numbers, n, n+1, n+2, ..., n+m-1." Sierpinski (p. 28). No proof is given in book.

%D W. Sierpinski, Pythagorean Triangles, Dover Publications, Mineola NY, 2003.

%H D. L. Vestal, <a href="http://www.maa.org/reviews/pythtriangles.html">Review of "Pythagorean Triangles"(Chapter 6) by W. Sierpinski</a>

%e a(4)=50 since 50, 51, 52 and 53 is the first occurrence of 4 consecutive integers which are lengths of hypotenuses of Pythagorean triangles.

%t lmt = 5*10^6; hyp = {5}; Do[ mn = m^2 + n^2; hyp = Join[hyp, Table[k*mn, {k, Floor[lmt/mn]}]]; hyp = Union[hyp], {n, 2, 1150}, {m, Min[n - 1, Floor[ Sqrt[ lmt - n^2]]]}]; f[n_] := Block[{k = 1}, While[phk[[k]] + n - 1 != phk[[k + n - 1]], k++ ]; phk[[k]]]; Do[ Print[ f[n]], {n, 33} (* _Robert G. Wilson v_, Nov 10 2004 *)

%Y See A099799 for another version.

%K nonn

%O 1,1

%A _Charlie Marion_, Nov 05 2004

%E More terms from _Ray Chandler_ and _Robert G. Wilson v_, Nov 10 2004